MATH 2203 ŒExam 4 (Version 1) Solutions

[Pages:5]S. F. Ellermeyer

MATH 2203 ?Exam 4 (Version 1) Solutions November 9, 2005 Name

1. Estimate the volume of the solid that lies below the surface z = x + 2y2 and above the rectangle R = [0; 2] [0; 4]. Use a Riemann sum with m = n = 2 and choose the sample points to be lower right corners. Repeat this problem using a midpoint Riemann sum.

Solution: When we subdivide the rectangle R into four pieces, the lower right corners of the subrectangles are at the points (1; 0), (2; 0), (1; 2), and (2; 2). Also, the area of each subrectangle is A = 2.

y4 3.5 3 2.5 2 1.5 1 0.5 0 0

0.5 1

1.5 2 x

The Riemann sum approximation of the volume of the solid described in this problem is

f (1; 0) A + f (2; 0) A + f (1; 2) A + f (2; 2) A where f is the function f (x; y) = x + 2y2. Thus the approximation is

(1) (2) + (2) (2) + (9) (2) + (10) (2) = 44.

When we use midpoints to do the approximation, we obtain

f (0:5; 1) A + f (1:5; 1) A + f (0:5; 3) A + f (1:5; 3) A = (2:5) (2) + (3:5) (2) + (18:5) (2) + (19:5) (2) = 88.

2. Use an iterated integral to compute the exact volume of the solid described in problem 1.

1

Solution: The exact volume is

ZZ

Z 2Z 4

f (x; y) dA =

x + 2y2 dy dx

00

R

= Z 2 xy + 2 y3 y=4 dx

Z0 2

3 y=0 128

=

4x +

dx

0

3

=

2x2

+

128 x

x=2

3 x=0

256 =8+ 3

= 280 3

= 93:3.

3. Find the volume of the solid that lies under the surface z = xy and above the triangular region with vertices at the points (1; 1), (4; 1), and (1; 2).

The answer to this problem is 31=8. However, as you will see, the calculations are very tedious. You will get full credit if you correctly set up the integral needed to compute the volume and perform at least the "inner"integration that needs to be done. If you get that far, then go on to the next problem and come back to this one if you have time left at the end of the test.

Solution: The triangular region, which we will call D, is pictured below. We will view it as a Type I region.

2

The volume of the solid that is described is

ZZ

Z Z 4

1 3

x+

7 3

xy dA =

xy dy dx

11

D

= Z 4 1 xy2 y=

1 3

x+

7 3

dx

1

Z

4

2 "

1

y=1

1

# 72 1

=

x x+

x dx

12

33

2

Z4 =

1 x(

x + 7)2

1 x dx

Z1 4 =

18 1

x3

2 14x2 + 49x

9 x dx

=

1

1

Z

18

4

x3

18 14x2 + 40x dx

18 1

= 1 1 x4 18 4

14 x3 + 20x2 x=4

3

x=1

1

896

=

64

+ 320

18

3

1 14 + 20

43

1

882 1

= 18 364 3 4

31 = 8.

3

4. Let D be the region pictured below.

Use polar coordinates to evaluate

ZZ xy dA.

D

Solution:

ZZ

Z Z3

xy dA =

r cos ( ) r sin ( ) r dr d

01

D

Z Z3

=

r3 sin ( ) cos ( ) dr d

0Z 1

1 =

4 Z0

r4 sin (

) cos (

)

r=3 r=1

d

1

=

80 sin ( ) cos ( ) d

40

= 20 1 sin2 ( ) =

2

=0

= 0.

5. Find the mass and center of mass of the rectangular lamina R = [0; 1] [0; 2] with density function (x; y) = y. Can you predict the location of the center of mass before actually doing the computations?

Solution:

4

The mass of the lamina is

ZZ

m=

(x; y) dA

Z

R 1

Z

2

=

y dy dx

=

0Z 1

0 1

2

Z

0 1

y2

y=2 y=0

dx

= 2 dx

0

= 2.

The moment with respect to the x axis is ZZ

Mx = y (x; y) dA

Z

R 1

Z

2

=

y2 dy dx

=

0Z 1

0 1

3 8

Z0

1

y3

y=2 y=0

dx

=

dx

30

8 = 3.

The moment with respect to the y axis is ZZ

My = x (x; y) dA

Z

R 1

Z

2

=

xy dy dx

=

0Z 1

0 1

2

Z

0 1

xy2

y=2 y=0

dx

= 2 x dx

Z

0 1

=

x2

x=1 x=0

dx

0

= 1.

The center of mass is

(x; y) =

My ; Mx

=

14 ;

.

mm

23

5

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